Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $y = \dfrac{-2}{6(5r + 3)} \div \dfrac{3r}{r(5r + 3)} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{-2}{6(5r + 3)} \times \dfrac{r(5r + 3)}{3r} $ When multiplying fractions, we multiply the numerators and the denominators. $y = \dfrac{ -2 \times r(5r + 3) } { 6(5r + 3) \times 3r } $ $ y = \dfrac{-2r(5r + 3)}{18r(5r + 3)} $ We can cancel the $5r + 3$ so long as $5r + 3 \neq 0$ Therefore $r \neq -\dfrac{3}{5}$ $y = \dfrac{-2r \cancel{(5r + 3})}{18r \cancel{(5r + 3)}} = -\dfrac{2r}{18r} = -\dfrac{1}{9} $